Q. If 3 F of electricity is passed through the solutions of ${AgNO}_{3},\ {CuSO}_{4}\ and\ {AuCl}_{3},$ the molar ratio of the cations deposited at the cathodes will be
A) 1:1:1
B) 1:2:3
C) 3:2:1
D) 6:3:2
Explanation
(d)
The increasing order of deposition of cations at the cathode is
${Cu}^{2 +} < {Ag}^{+} < {Au}^{3 +}$
$\ \ \ \ \ \ \ E\ \propto \ Z$
$\ \ \ {Ag}^{+}\ + \ e^{-}\ \rightarrow \ Ag$
$\ {Cu}^{2 +}\ + \ {2e}^{-}\ \rightarrow \ Cu$
$\ {Au}^{3 +}\ + \ {3e}^{-}\ \rightarrow \ Au$
3 Faradays liberate 1 mole of Au, 3 moles of Ag and 3/2 moles of Cu. Thus, molar ratio of Ag:Cu: Au is 3:3/ 2:1 or 6:3:2.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.